I have two questions:
(1) What is a discrete category functor?
(2) Malice Vidrine's answer to Functor over sets of functions implies that there is a relationship between the exponential object $B^A$ and "the (discrete) functor category of $A,B$ thought of as categories. What does he mean?
My questions relate to confusions understanding the reply to this question: What is the carrier set functor?
Given a collection of objects $C$ (usually a set in this context), there is an "obvious" category that has objects $C$: it's the category whose only morphisms are the identity morphisms with obvious composition, which is often called the discrete category associated to $C$.
In the answer to your previous question, this category was denoted $DC$. It's obvious that any map $f: C\to E$ can be thought of as a functor $Df: DC \to DE$ that sends the object $c\in C$ to $f(c)$, and the arrow $id_c$ to $id_{f(c)}$. Since in both $DC, DE$ there are only trivial compositions, these things fill out to be a functor $D: Set \to \mathbf{Cat}$, where $\mathbf{Cat}$ is the category of small categories and functors between them. This is the "discrete category" functor, because it sends sets to the associated discrete category.
So in your previous question, the notation $Set^S$ can be equivalently seen as : the category of functors $DS \to Set$ where $DS$ is as above, or as the category of $S$-indexed families. Generally, for sets $A,B$, $B^A$ can be seen as the category of functors $DA\to DB$ because such a functor is precisely a map $f: A\to B$, i.e. the assignement $D: Hom_{Set}(A,B) \to Hom_{\mathbf{Cat}}(DA,DB)$ is full (and faithful): $D$ is a full (and faithful) functor.
What's interesting is that this last analogy between $B^A$ and $Hom_{\mathbf{Cat}}(DA,DB)$, is that $D(B^A) \cong [DA,DB]$ where $[DA,DB]$ is the functor category. That's because if you have $Df, Dg: DA\to DB$ two functors, and $\tau: Df\to Dg$ a natural transformation, then for any $a\in Ob(DA)=A$, $\tau_a: f(a) \to g(a)$ is a morphism in $DB$, which implies $g(a)= f(a)$. Therefore if there is a natural transformation (or a $[DA,DB]$-morphism) $Df \to Dg$, we get $f=g$ and this natural transformation is the identity, hence $Df= Dg$, that is, $[DA,DB]$ is the discrete category associated to $B^A$.
